JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 14)
Let the lengths of intercepts on x-axis and y-axis made by the circle
x2 + y2 + ax + 2ay + c = 0, (a < 0) be 2$${\sqrt 2 }$$ and 2$${\sqrt 5 }$$, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is equal to :
x2 + y2 + ax + 2ay + c = 0, (a < 0) be 2$${\sqrt 2 }$$ and 2$${\sqrt 5 }$$, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is equal to :
$${\sqrt {10} }$$
$${\sqrt {6} }$$
$${\sqrt {11} }$$
$${\sqrt {7} }$$
Explanation
$$2\sqrt {{{{a^2}} \over 4} - c} = 2\sqrt 2 $$
$$\sqrt {{a^2} - 4c} = 2\sqrt 2 $$
$${a^2} - 4c = 8$$ .... (1)
$$2\sqrt {{a^2} - c} = 2\sqrt 5 $$
$${a^2} - c = 5$$ .... (2)
$$(2) - (1)$$
$$3c = - 3a \Rightarrow c = - 1$$
$${a^2} = 4 \Rightarrow a = - 2$$ (Given a < 0)
Equation of circle
$${x^2} + {y^2} - 2x - 4y - 1 = 0$$
Equation of tangent which is perpendicular to the line x + 2y = 0 is
$$2x - y + \lambda = 0$$
$$ \therefore $$ p = r
$$\left| {{{2 - 2 + \lambda } \over {\sqrt 5 }}} \right| = \sqrt 6 $$
$$ \Rightarrow \lambda = \pm \sqrt {30} $$
$$ \therefore $$ Tangent $$2x - y \pm \sqrt {30} = 0$$
Distance from origin = $${{\sqrt {30} } \over {\sqrt 5 }} = \sqrt 6 $$
$$\sqrt {{a^2} - 4c} = 2\sqrt 2 $$
$${a^2} - 4c = 8$$ .... (1)
$$2\sqrt {{a^2} - c} = 2\sqrt 5 $$
$${a^2} - c = 5$$ .... (2)
$$(2) - (1)$$
$$3c = - 3a \Rightarrow c = - 1$$
$${a^2} = 4 \Rightarrow a = - 2$$ (Given a < 0)
Equation of circle
$${x^2} + {y^2} - 2x - 4y - 1 = 0$$
Equation of tangent which is perpendicular to the line x + 2y = 0 is
$$2x - y + \lambda = 0$$
$$ \therefore $$ p = r
$$\left| {{{2 - 2 + \lambda } \over {\sqrt 5 }}} \right| = \sqrt 6 $$
$$ \Rightarrow \lambda = \pm \sqrt {30} $$
$$ \therefore $$ Tangent $$2x - y \pm \sqrt {30} = 0$$
Distance from origin = $${{\sqrt {30} } \over {\sqrt 5 }} = \sqrt 6 $$
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